Question

Information from the American Institute of Insurance indicates the mean amount of life insurance per household in the United States is $118,000. This distribution follows the normal distribution with a standard deviation of $32,000.a. If we select a random sample of 56 households, what is the standard error of the mean? (Round your answer to the nearest whole number.) Standard error of the mean b. What is the expected shape of the distribution of the sample mean? Sample mean (Click to select)Unknown.Not normal, the standard deviation is unknown.Normal.Uniformc. What is the likelihood of selecting a sample with a mean of at least $125,000? (Round z value to 2 decimal places and final answer to 4 decimal places.) Probability d. What is the likelihood of selecting a sample with a mean of more than $108,000? (Round z value to 2 decimal places and final answer to 4 decimal places.) Probability e. Find the likelihood of selecting a sample with a mean of more than $108,000 but less than $125,000.(Round z value to 2 decimal places and final answer to 4 decimal places.) Probability

Answer 1

Explanation:

Hello!

The variable of interest is

X: the amount of life insurance per household in the United States.

This variable follows a normal distribution with mean μ= $188000 and standard deviation σ=$32000

a.

If a sample n=56The sampling distribution is X[bar]~N(μ;σ²/n)

The standard error fro the sample mean is σ/√n= 32000/√56= 4276.1798≅ $4276.18

b.

If the population of origin has an exact normal distribution and from this population, many random samples are taken and their mean calculated, the sample mean will be also a random variable with the same type of distribution as the original variable. In this case, since the variable of interest has a normal distribution, then the sample mean will also have a normal distribution.

c.

P(X[bar]≥$125000) = 1 - P(X[bar]<$125000)= 1 - P(Z<(125000-118000)/4276.18)= 1 - P(Z<1.64)= 1 - 0.94950= 0.0505

d.

P(X[bar]>$108000)= 1 - P(X[bar]≤$108000)= 1 - P(Z≤(108000-118000)/4276.18)= 1 - P(Z≤-2.34)= 1 - 0.00964= 0.99036

e.

P(108000≤X[bar]≤12500)= P(X[bar]≤125000) - P(X[bar]≤108000)= P(Z≤1.64) - P(Z≤-2.34)= 0.94950 - 0.00964= 0.93986

I hope this helps!